Wheatstone Bridge & Metre Bridge question from MHT CET 2024 (11 May, Shift 1)

Answer: (b)

1. Redraw: the two $6\ \Omega$ go from the left terminal to nodes X and Y; the two $4\ \Omega$ go from X and Y to the right terminal; $10\ \Omega$ joins X and Y.
2. $\dfrac{6}{4} = \dfrac{6}{4}$, so the bridge is balanced and $10\ \Omega$ carries no current.
3. $R_{eq} = (6+4) \parallel (6+4) = 5\ \Omega$
4. $I = \dfrac{12}{5} = 2.4$ A
5. So the answer is (b) $2.4$ A.